$f(x, y) = \left( x^3y, - xy^3 \right)$ $\text{div}(f) = $
Explanation: The formula for divergence in two dimensions is $\text{div}(f) = \dfrac{\partial P}{\partial x} + \dfrac{\partial Q}{\partial y}$, where $P$ is the $x$ -component of $f$ and $Q$ is the $y$ -component. Let's differentiate! $\begin{aligned} \dfrac{\partial P}{\partial x} &= \dfrac{\partial}{\partial x} \left[ x^3y \right] \\ \\ &= 3x^2y \\ \\ \dfrac{\partial Q}{\partial y} &= \dfrac{\partial}{\partial y} \left[ -y^3x \right] \\ \\ &= -3xy^2 \end{aligned}$ Adding the two partial derivatives, $\text{div}(f) = 3x^2y - 3xy^2$.